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Evaluate the Following Limits: Lim X → 0 2 Sin X − Sin 2 X X 3 - Mathematics

Evaluate the following limits: 

\[\lim_{x \to 0} \frac{2\sin x - \sin2x}{x^3}\] 

 

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Solution

\[\lim_{x \to 0} \frac{2\sin x - \sin2x}{x^3}\]
\[ = \lim_{x \to 0} \frac{2\sin x - 2\sin x\cos x}{x^3}\]
\[ = \lim_{x \to 0} \frac{2\sin x\left( 1 - \cos x \right)}{x^3}\]
\[ = \lim_{x \to 0} \frac{2\sin x \times 2 \sin^2 \frac{x}{2}}{x^3}\] 

\[= \lim_{x \to 0} \frac{\sin x \times \sin^2 \frac{x}{2}}{x \times \frac{x^2}{4}}\]
\[ = \lim_{x \to 0} \frac{\sin x}{x} \times \left( \lim_{x \to 0} \frac{\sin\frac{x}{2}}{\frac{x}{2}} \right)^2 \]
\[ = 1 \times 1 \left( \lim_\theta \to 0 \frac{\sin\theta}{\theta} = 1 \right)\]
\[ = 1\]

 

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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 29 Limits
Exercise 29.7 | Q 51 | Page 51
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